If #cos(t) = −7/19# and tan(t) < 0, find sin(t) and cos(-t). ?

Give the answer as a fraction with a square root. Hint: Draw a triangle (perhaps within the unit circle) containing an angle with sides for the given trigonometric function.

2 Answers
May 2, 2018

#sin(t)=(2sqrt78)/361#
#cos(-t)=-7/19#

Explanation:

#tan(t)<0# simply gives us quadrant information - the tangent function is negative in quadrants II and IV, while the cosine function is negative in quadrants II and III. Therefore #t# must occur in quadrant II.

To find #sin(t)#, you can use the identity #sin^2t+cos^2t=1#. Inputting knowns:
#sin^2t+(-7/19)^2=1# Simplify:
#sin^2t=312/361#
#sin(t)=(sqrt312)/361# or #(2sqrt78)/361#
We know this is positive because sine is positive in QII.

Because cosine is an even function, #cos(-t)=cos(t)#, so #cos(-t)=-7/19#

May 2, 2018

#sint=(2sqrt78)/19and cos(-t)=-7/19#

Explanation:

Here,

#cost=-7/19=> color(blue)(II^(nd)Quadrant )orIII^(rd)Quadrant...to(1)#

#tant < 0=> color(blue)(II^(nd) Quadrant) or IV^(th)Quadrant...to(2)#

From #(1) and (2)#,we can say that

#pi/2<= t <= pi to color(blue)(II^(nd) Quadrant)=>(sine) > 0 #

So,

#(i).sin(t)=+sqrt(1-cos^2t)=sqrt(1-49/361)=sqrt312/19= (2sqrt78)/19#

We know that,

#"cosine is "color(red)"even function"=>cos(-theta)=costheta#

#(ii).cos(-t)=cost=-7/19#

Note:

#pi/2<= t <= pi =>-pi/2 >= (-t) >= -pi#

#i.e.-pi <= (-t) <= -pi/2=>III^(rd) Quadrant=># cosine #< 0#